Question

A stone of mass $1.3 \, kg$ is being rotated in a horizontal plane as a conical pendulum with the help of a $140 \, cm$ long aluminium wire of cross-sectional area $1 \, mm^{2}$ . The wire makes an angle $\theta =75^\circ $ with the vertical. What is the increment in the length (in $mm$ ) of the wire? [Young's modulus of aluminium $Y_{Al}=7\times 10^{10} \, \text{N m}^{\text{-2}}$ , $sin75^\circ \approx 0.97$ , $cos75^\circ \approx 0.26$ , $g=10 \, m \, s^{- 2}$ ]

Answer 1

The figure below shows the stone being rotated as a conical pendulum such that the wire makes an angle of $75^\circ $ with the vertical. Let us assume that the radius $OA$ of the circle is equal to $r$

For vertical equilibrium of the stone
$Tcos75^\circ =Mg$
$T=\frac{M g}{cos 75 ^\circ }$
The elongation in the wire is
$\Delta L=\frac{T L}{A Y}=\frac{M g L}{A Y cos 75 ^\circ }$
$\Rightarrow \Delta L=\frac{1 . 3 \times 10 \times 1 . 4}{10^{- 6} \times 7 \times 10^{10} \times 0 . 26}=10^{- 3}m=1mm$